Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
62 views
in Physics by (91.5k points)
closed by
A `0.1 m` long conductor carrying a current of `50 A` is perpendicular to a magentic field of `1.25 mT`. The mechanical power to move the conductor with a speed of `1ms^(-1)` is
A. `62.5 mW`
B. `625 mW`
C. `6.25 mW`
D. `12.5 mW`

1 Answer

0 votes
by (91.6k points)
selected by
 
Best answer
Correct Answer - C
Here, l = 0.1m, v = 1 `ms^(-1)`
I = 50 A and B = 1.25 mT = `1.25xx10^(-3)T`
The induced emf is, `epsilon=Blv`
The mechanical power is,
`P=epsilonI=BlvI=1.25xx10^(-3)xx0.1xx1xx50`
`=6.25xx10^(-3)W=6.25mW`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...